Acceleration of a horizontal system with friction

  • #1

Homework Statement



Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (Fig. P5.43). Suppose F = 68.0 N, m1 = 12.0 kg, m2 = 18.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.100. (a) Draw a free-body diagram for each block. (b) Determine the tension T and the magnitude of the acceleration of the system.



Homework Equations



F = ma - fk

fk = μkmg

The Attempt at a Solution



68.0 N = (12.0 kg + 18.0 kg)*a – (0.100)(12.0 kg)(9.8 m/s2) – (0.100)(18.0 kg)(9.8 m/s2)

a = [68.0 N + (0.100)(12.0 kg)(9.8 m/s2) + (0.100)(18.0 kg)(9.8 m/s2)]/(12.0 kg + 18.0 kg)

a = 3.25 m/s2

T = ma - fk

T = (12.0 kg)( 3.25 m/s2) - (0.100)(12.0 kg)(9.8 m/s2)

T = 27.2 N

The tension of T is correct according to the answer key in the book but my acceleration is not. I cannot understand this since the acceleration I have is what I used to determine T. The book states a = 1.29 m/s2

The only thing I can think of is that since friction is involved that there must be a vertical acceleration that must be taken into account then I could square each acceleration of the x and y direction, add them and take the square root. Am I on the right track and how would I determine the acceleration in the y direction?
 

Answers and Replies

  • #2
Doc Al
Mentor
45,011
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68.0 N = (12.0 kg + 18.0 kg)*a – (0.100)(12.0 kg)(9.8 m/s2) – (0.100)(18.0 kg)(9.8 m/s2)
Newton's 2nd law tells us:
ΣF = ma

The net force will be the applied force minus the friction forces.
 
  • #3
Newton's 2nd law tells us:
ΣF = ma

The net force will be the applied force minus the friction forces.
I get that. That is how I setup the problem. I have the net force, and the total mass, and the friction forces. What I didn't have was the acceleration. So with basic algebra I calculated what I thought was acceleration. Then to determine the force on the string connecting the two blocks I used the acceleration I found times the mass of second block minus the friction of just that block. So why isn't that acceleration correct?
 
  • #4
Doc Al
Mentor
45,011
1,288
I get that. That is how I setup the problem. I have the net force, and the total mass, and the friction forces. What I didn't have was the acceleration. So with basic algebra I calculated what I thought was acceleration.
Your equation (that I quoted) is incorrect. Find the net force (on the left hand side) and set it equal to ma.
Then to determine the force on the string connecting the two blocks I used the acceleration I found times the mass of second block minus the friction of just that block. So why isn't that acceleration correct?
That reflects a similar error. Again, set the net force equal to ma, then solve for the tension.
 
Last edited:
  • #5
68.0 N – (0.100)(12.0 kg)(9.8 m/s^2) – (0.100)(18.0 kg)(9.8 m/s^2) = (12.0 kg + 18.0 kg)*a

I must have misinterpreted this in class or from the book. So it's not F = ma - fk but instead F - fk = ma

So is it just a coincidence that the Tension worked out before? I get the same answer with the a = 1.29 when I use the formula F - fk = ma
 
  • #6
Doc Al
Mentor
45,011
1,288
68.0 N – (0.100)(12.0 kg)(9.8 m/s^2) – (0.100)(18.0 kg)(9.8 m/s^2) = (12.0 kg + 18.0 kg)*a
There you go.

I must have misinterpreted this in class or from the book. So it's not F = ma - fk but instead F - fk = ma
Right. It's best to think in terms of Fnet = ΣF = ma. (All forces go under ΣF.)

So is it just a coincidence that the Tension worked out before?
Maybe you're just lucky? :wink: (I didn't check, but I suppose the two errors canceled themselves. Don't count on it!)
 
  • #7
Well I had a similar issue in class when doing a frictionless problem dealing with two objects, one pushing against the other. In the example first given they were level, and in the problem we work they were on an incline. Before the teacher started I worked it out exactly as the example getting the right answer, I checked after doing the work. She then did the example using the the angle given to compensate for the force or gravity and the normal. I understand why she did it that way but when I asked why it worked out right without using the angle, I asked if it was a coincidence, she didn't have an answer except to say, "If you do it that way you won't get credit." Good enough for me I guess, but sometimes I just really want to know. And what you said at least makes sense, they likely did cancel out since I made same error twice by using the formula incorrectly. Thanks so much for the help, I will be posting a lot more now that I know I can get exactly the kind of assistance I need, not the answer but how to get it yourself. ;-)
 

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